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Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Overtime Rule in Football Before the overtime rule in the National Football League was changed in 2011, among 460 overtime games, 252 were won by the team that won the coin toss at the beginning of overtime. Using a 0.05 significance level, test the claim that the coin toss is fair in the sense that neither team has an advantage by winning it. Does the coin toss appear to be fair?

Short Answer

Expert verified

Null hypothesis: The proportion of overtime wins that arewon by the team that has won the coin toss is equal to 0.5.

Alternative hypothesis: The proportion of overtime wins that arewon by the team that has won the coin toss is not equal to 0.5.

Test statistic: 2.05

Critical value: 1.96

P-value: 0.0404

The null hypothesis is rejected.

There is enough evidence to reject the claim that the proportion of games that arewon by the team that has won the coin toss is equal to 0.5.

The coin toss does not appear to be fair.

Step by step solution

01

Given information

Out of 460 overtime games, 252 of them arewon by the team that has won the coin toss. It is claimed that the coin toss is fair, and an equal number of teams who win/do not win the coin toss go on to win the game.

02

Hypotheses

The null hypothesis is written as follows.

The proportion of overtime wins that arewon by the team that has won the coin toss is equal to 0.5.

\({H_0}:p = 0.5\).

The alternative hypothesis is written as follows.

The proportion of overtime wins that arewon by the team that has won the coin toss is not equal to 0.5.

\[{H_1}:p \ne 0.5\].

The test is two-tailed.

03

Sample size, sample proportion,and population proportion

The sample size is equal to n=460.

The sample proportion of overtime wins that arewon by the team that has won the coin toss is computed below.

\[\begin{array}{c}\hat p = \frac{{252}}{{460}}\\ = 0.548\end{array}\].

The population proportion of games that arewon by the team that has won the coin toss is equal to 0.5.

04

Test statistic

The value of the test statistic is computed below.

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.548 - 0.5}}{{\sqrt {\frac{{0.5\left( {1 - 0.5} \right)}}{{460}}} }}\\ = 2.05\end{array}\).

Thus, z=2.05.

05

Critical value and p-value

Referring to the standard normal table, the critical value of z at\(\alpha = 0.05\)for a two-tailed test is equal to 1.96.

Referring to the standard normal table, the p-value for the test statistic value of 2.05 is equal to 0.0404.

Asthe p-value is less than 0.05, the null hypothesis is rejected.

06

Conclusion of the test

There is enough evidence to reject the claim that the proportion of games that arewon by the team that has won the coin toss is equal to 0.5.

This concludes that the coin toss is not fair, and there is a greater chance of winning the game after winning the coin toss.

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Most popular questions from this chapter

Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Survey Return Rate In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 5000 subjects randomly selected from an online group involved with ears. 717 surveys were returned. Use a 0.01 significance level to test the claim that the return rate is less than 15%.

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

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Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Got a Minute? Students of the author estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Use a 0.05 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute?

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In Exercises 1–4, use these results from a USA Today survey in which 510 people chose to respond to this question that was posted on the USA Today website: “Should Americans replace passwords with biometric security (fingerprints, etc)?” Among the respondents, 53% said “yes.” We want to test the claim that more than half of the population believes that passwords should be replaced with biometric security.

Requirements and Conclusions

a. Are any of the three requirements violated? Can the methods of this section be used to test the claim?

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c. Another memory trick commonly used is this: “If the P is high, the null will fly.” Given that a hypothesis test never results in a conclusion of proving or supporting a null hypothesis, how is this memory trick misleading?

d. Common significance levels are 0.01 and 0.05. Why would it be unwise to use a significance level with a number like 0.0483?

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